Question

Cho x thuộc Z, chứng minh rằng \(x^{200}+x^{100}+1⋮x^4+x^2+1\)

Answer 1

 \(A=x^{200}+x^{100}+1\)

    \(=x^{200}-x^2+x^{100}-x^4+x^4+x^2+1\)

    \(=x^2\left(x^{198}-1\right)+x^4\left(x^{96}-1\right)+\left(x^4+x^2+1\right)\)

    \(=x^2\left(x^{^6}-1\right).A+x^4\left(x^6-1\right).B+x^4+x^2+1\)

\(x^6-1=\left(x^3-1\right)\left(x^3+1\right)=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)=\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)\)

Vậy \(A⋮\left(x^4+x^2+1\right)\)